Question

please tell the rigt answer​

Answer 1

Answer:

2

Step-by-step explanation:

2 sin 52/cos 38-tan 70/cot 20+sin 63. sec 27

2 sin 52/sin(90-38)-tan 70/tan(90-20)+sin 63/sin(90-27)

2 sin 52/sin 52-tan 70/tan 70 +sin 63/sin 63

2-1+1=2

Answer 2

$\mathsf{Given : \dfrac{2Sin52^{\circ}}{Cos38^{\circ}} - \dfrac{Tan70^{\circ}}{Cot20^{\circ}} + Sin63^{\circ}Sec27^{\circ}}$

We know that :

$\bigstar \ \ \boxed{\mathsf{Sin(90 - \theta) = Cos\theta}}$

$\bigstar \ \ \boxed{\mathsf{Tan(90 - \theta) = Cot\theta}}$

Now, Consider : Sin52°

It can be written as : Sin(90° - 38°) = Cos38°

$:\implies$ Sin52° = Cos38°

Consider : Tan70°

It can be written as Tan(90° - 20°) = Cot20°

$:\implies$ Tan70° = Cot20°

Substituting the above values in the question, We get :

$\mathsf{\implies \dfrac{2Cos38^{\circ}}{Cos38^{\circ}} - \dfrac{Cot20^{\circ}}{Cot20^{\circ}} + Sin63^{\circ}Sec27^{\circ}}$

$\mathsf{\implies 2 - 1 + Sin63^{\circ}Sec27^{\circ}}$

Now, Consider : Sin63°

It can be written as : Sin(90° - 27°) = Cos27°

$:\implies$ Sin63° = Cos27°

$\mathsf{\implies 2 - 1 + Cos27^{\circ}Sec27^{\circ}}$

We know that : [Cosθ . Secθ] = 1

$\mathsf{\implies 2 - 1 + 1}$

$\mathsf{\implies 2}$